Question

An electron is shot through a varying electrical field. Initially the electron is moving in the x-direction with a velocity V_x = 100 m/s.The electron enters the field when it passes the origin. The field varies with time, causing an accele(ration of the electron that varies in time:

a(t) = (-20 m/s^2 - 10 m/s^3*t)

Already calculated information:

V_y = -20t - 5t^2 (velocity - y-axis)

r_x = 100*t (position - x-axis)

r_y = -10t^2 - 5/3*t^3 (position - y-axis)

The field is acting inside a box of length L = 2m, and the electron is inside the field for t = 1/50 sek (or 0.02 sek)

What is the displacement in the y-direction when the electron leaves the box?

Find the angle the velocity vector forms with the horizontal as the electron leaves the box.

Answer 1

The vertical position is given by: 13 = -1062- For t = 0.02 s Then: 1.; = –10(0.02) - 5 (0.02) ry = -4.01 x 10-3 m The horizontal component of the velocity is: Vx = 100" The vertical component of the velocity at t = 0.02 s is. Vy = –20(0.02) – 5(0.02)2 = -0.402" The angle is given by: = tan-1 0.402 100 = -0.23